Let $k$ be a positive real number. The square with vertices (k, 0), (0, k), (-k, 0), and (0, -k) is plotted in the coordinate plane.
Find conditions on a > 0 and b > 0 such that the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\] is contained inside the square (and tangent to all of its sides). HINTS: Suppose that the line x+y=k is tangent to the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\]. Algebraically, what can we say about the solutions? In particular, the number of solutions?
+26367 Let \(k\) be a positive real number. The square with vertices \((k, 0)\), \((0, k)\), \((-k, 0)\), and \((0, -k)\) is plotted in the coordinate plane.
Find conditions on \(a > 0\) and \(b > 0\) such that the ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) is contained inside the square (and tangent to all of its sides).
I Suppose that the line \(\mathbf{y=x+ k}\) is tangent to the ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\).
My answer see here: https://web2.0calc.com/questions/question-about-conics-pls-help#r3
@heureka, what do $x_t$ and $y_t$ mean in your solution? Are they bases from logs?\(\)
+26367 @heureka, what do $x_t$ and $y_t$ mean in your solution? Are they bases from logs?
no
\((x_t,y_t)\) are the coordinates of the tanget point P at the ellipse on my line \(y = x+k\). See my link and the graph.
The tanget point of the circle is \((-10,10)\), so \(x_t=-10\) and \(y_t =10\).
